2 Answers
2

In the paper "On non-modular n-distributive lattices: Lattices of convex sets" (Acta Sci. Math. (Szeged) 52 (1987), 35-45) A. Hunh proves that the lattice of convex sets in $\mathbb R^n$, $Co(\mathbb R ^n)$, is $n+1$-distributive but not $n$-distributive. This means that the lattice of convex sets in $\mathbb R^n $ is only distributive for $n=0$, as the example by François G. Dorais shows.

Here we define a lattice $L$ to be $n$-distributive if for any $x,y_0,\dots,y_n$ the following identity holds:
$$x\wedge \bigvee_{i=0}^n y_i =\bigvee_{i=0}^n (x \wedge \bigvee_{i\neq j} y_i).$$
This nice result shows that the dimension of a Euclidean space has a lattice theoretical characterization. The proof relies on Caratheodory's theorem. The same holds for the dual of $Co(\mathbb{R}^n)$, it is $n+1$-distributive but not $n$-distributive, and the proof uses Helly's theorem.

I googled "lattice of convex sets"+distributive and the second link led me to the abstract for the paper: "Geometric Condition for Local Finiteness of a Lattice of Convex Sets"
in Mathematica Moravica, Vol. 1 (1997), 35–40 by Matt Insall, which contains the following sentence:

We give elementary examples which establish the following facts: the lattice of convex subsets of a Hilbert space is not locally finite, it is not modular (hence not distributive), and locally finite lattices of closed convex sets in any Hilbert space have very restrictive geometric arrangements of their members.

This means that the answer to your question is no, since $\mathbb{R}^2$ is certainly a Hilbert space.

Yeah, having glanced at the paper, it appears that what they actually prove is explicitly in the plane, but apparently they were more concerned with the result for general nonzero complex Hilbert spaces. However, as François G. Dorais's comment shows, this is overkill for this particular question.
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Jonas MeyerJan 25 '10 at 6:24